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Match the following List-I with List -II
List-I
List-II
(I)
The shortest distance between the parabola $2 y^{2} = 2 x - 1$ and $2 x^{2} = 2 y - 1$ is
(P)
$\frac{2 \sqrt{2}}{3}$
(II)
The triangle formed by the tangent to the parabola $y = x^{2}$ at the point whose abscissa $(x_{0} \in [1, 2]),$ is the y-axis and the straight line $y = x_{0}^{2}$ has the greatest area if $x_{0}$ is equal to
(Q)
$\frac{1}{4}$
(III)
An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is $\frac{2}{3}$ , then the eccentricity of the ellipse is
(R)
$\frac{1}{2 \sqrt{2}}$
(IV)
If $f (x) = x^{3} + a x^{2} + b x + c,$ (a,b,c are rational numbers) and 2 roots of $f (x) = 0$ are eccentricities of a parabola and a rectangular hyperbola, then $\frac{1}{a b c}$ equals
(S)
$\frac{1}{\sqrt{2}}$

(T)
2

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$(I) \to (S); (I I) \to (T); (I I I) \to (Q); (I V) \to (P)$

$(I) \to (R); (I I) \to (Q); (I I I) \to (S); (I V) \to (T)$

$(I) \to (S); (I I) \to (Q); (I I I) \to (R); (I V) \to (T)$

$(I) \to (R); (I I) \to (T); (I I I) \to (P); (I V) \to (Q)$

answer is D.

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Detailed Solution

(I) The given parabolas are symmetric about the line y=x
(II) Area of the given triangle $= x_{0}^{3}$
(III) Probability (required) $= \frac{π a^{2} - π a b}{π a^{2}} = \frac{2}{3} \Rightarrow a = 3 b$
(IV) Here, roots are $1, \sqrt{2}, - \sqrt{2}$

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